If |vec{a}| = sqrt{26},|vec{b}| = 7,and |vec{ | vedclass

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(D) We are given $|\vec{a}| = \sqrt{26}$,$|\vec{b}| = 7$,and $|\vec{a} 	imes \vec{b}| = 35$.Using the formula for the cross product: $|\vec{a} 	imes

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$7$

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(D) We are given $|\vec{a}| = \sqrt{26}$,$|\vec{b}| = 7$,and $|\vec{a} 	imes \vec{b}| = 35$.Using the formula for the cross product: $|\vec{a} 	imes \vec{b}| = |\vec{a}| |\vec{b}| \sin 	heta = 35$.Substituting the values: $\sqrt{26} 	imes 7 	imes \sin 	heta = 35$.$\sin 	heta = \frac{35}{7 \sqrt{26}} = \frac{5}{\sqrt{26}}$.Since $\sin^2 	heta + \cos^2 	heta = 1$,we have $\cos^2 	heta = 1 - \sin^2 	heta = 1 - \frac{25}{26} = \frac{1}{26}$.Thus,$|\cos 	heta| = \frac{1}{\sqrt{26}}$.The dot product is given by $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos 	heta$.$\vec{a} \cdot \vec{b} = \sqrt{26} 	imes 7 	imes \left( \pm \frac{1}{\sqrt{26}} ight) = \pm 7$.Given the options,the correct value is $7$.

If $|\vec{a}| = \sqrt{26}$,$|\vec{b}| = 7$,and $|\vec{a} \times \vec{b}| = 35$,find $\vec{a} \cdot \vec{b}$.

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